# Composition of an entire function with the principal part of a Laurent series

$$\def\bbC{\mathbb{C}} \def\bbR{\mathbb{R}}$$For $$r,R\in\bbR$$, denote $$A(r,R)=\{z\in\bbC:r<|z| to the open annulus in $$\bbC$$ of inner and outer radii $$r$$ and $$R$$, respectively. Suppose that $$f(z)=\sum_{k=-\infty}^{+\infty}a_{k}z^{k}$$ is a convergent Laurent series in $$A(r,R)$$. We call $$\sum_{k\geq0}a_kz^k$$ and $$\sum_{k<0}a_kz^k$$ the Taylor part and principal part of the Laurent series, respectively. Let $$g$$ be an entire function such that $$g(0)=0$$. We have that $$g\circ f$$ is a holomorphic function in $$A(r,R)$$ and hence it has a convergent Laurent series development in this annulus. My question is: if the Taylor part of $$f$$ vanishes, then the Taylor part of $$g\circ f$$ vanishes as well?

A physicist approach tells me that yes, since if $$g(z)=\sum_{k=1}^{+\infty}b_kz^k$$, then we have $$g(f(z))=b_1f(z)+b_2f(z)^2+b_3f(z)^3+\cdots$$ and by expanding $$f$$ and “reordering the terms,” we must end up with a Laurent series with vanishing Taylor part.

The “Taylor part” $$\sum_{k\geq0}a_kz^k$$ of $$f$$ vanishes if and only if $$f(z) = \sum_{k<0}a_kz^k$$ in $$A(r, R)$$, and that happens if and only if $$f$$ is the restriction of a function $$F$$ which is holomorphic in $$|z| > r$$ with $$\lim_{z \to \infty} F(z) = 0$$.
In that case is $$g \circ f$$ the restriction of $$g \circ F$$, which is also holomorphic in $$|z| > r$$ with $$\lim_{z \to \infty} g(F(z)) = 0$$, so that the Taylor part of $$g \circ f$$ vanishes as well.
Okay, here is a rigorous proof. Write the Laurent series of the composition $$g(f(z))=\sum_{k=-\infty}^{+\infty}c_kz^k,\quad z\in A(r,R).$$ We want to show that $$c_k=0$$ for all $$k\geq 0$$. Fix $$\rho$$ between $$r$$ and $$R$$. For any $$k$$, we have \begin{align} 2\pi ic_k&=\int_{|z|=\rho}\frac{g(f(z))}{z^{k+1}}dz\\ &=\int_{|z|=\rho}\sum_{n=1}^{+\infty}b_k\frac{f(z)}{z^{k+1}}dz\\ &=\sum_{n=1}^{+\infty}b_k\int_{|z|=\rho}\frac{f(z)}{z^{k+1}}dz, \end{align} where we used the Fubini-Tonelli theorem, seeing the sum as an integral with the counting measure.
By the same argument, we have \begin{align} \int_{|z|=\rho}\frac{f(z)}{z^{k+1}}dz &=\int_{|z|=\rho}\sum_{m=-1}^{-\infty}a_m\frac{z^m}{z^{k+1}}dz\\ &=\sum_{m=-1}^{-\infty}a_m\int_{|z|=\rho}\frac{z^m}{z^{k+1}}dz. \end{align} Thus, if $$k\geq 0$$, the integral $$\int_{|z|=\rho}\frac{z^m}{z^{k+1}}dz$$ vanishes and so does $$c_k$$.